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NCERT Solutions · Class 11 Maths · Chapter 1

Chapter 1: Sets (Sets and Functions)

Q: Write in roster form: (i) A = {x : x an integer, −3 ≤ x < 7} (ii) B = {x : x a natural number less than 6} (iii) C = {x : x a two-digit natural number whose digits add to 8} (iv) D = {x : x a prime divisor of 60} (v) E = letters of TRIGONOMETRY (vi) F = letters of BETTER.

(i) {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6}; (ii) {1, 2, 3, 4, 5}; (iii) {17, 26, 35, 44, 53, 62, 71, 80}; (iv) {2, 3, 5}; (v) {T, R, I, G, O, N, M, E, Y}; (vi) {B, E, T, R}

Q: Write in set-builder form: (i) {3, 6, 9, 12} (ii) {2, 4, 8, 16, 32} (iii) {5, 25, 125, 625} (iv) {2, 4, 6, …} (v) {1, 4, 9, …, 100}.

(i) {x : x = 3n, n ∈ N, 1 ≤ n ≤ 4}; (ii) {x : x = 2ⁿ, n ∈ N, 1 ≤ n ≤ 5}; (iii) {x : x = 5ⁿ, n ∈ N, 1 ≤ n ≤ 4}; (iv) {x : x is an even natural number}; (v) {x : x = n², n ∈ N, 1 ≤ n ≤ 10}

Q: List the elements: (i) A = {x : x an odd natural number} (ii) B = {x : x an integer, −½ < x < 9/2} (iii) C = {x : x an integer, x² ≤ 4} (iv) D = letters of LOYAL (v) E = months not having 31 days (vi) F = consonants preceding k.

(i) {1, 3, 5, …}; (ii) {0, 1, 2, 3, 4}; (iii) {−2, −1, 0, 1, 2}; (iv) {L, O, Y, A}; (v) {February, April, June, September, November}; (vi) {b, c, d, f, g, h, j}

Q: Match: (i) {1, 2, 3, 6} (ii) {2, 3} (iii) {M, A, T, H, E, I, C, S} (iv) {1, 3, 5, 7, 9} with (a) {x : x a prime divisor of 6} (b) {x : x an odd natural number less than 10} (c) {x : x a natural number and divisor of 6} (d) {x : x a letter of MATHEMATICS}.

(i) (c); (ii) (a); (iii) (d); (iv) (b)

Step-by-step solutions to all exercises of Chapter 1, Sets (NCERT Class 11 Maths, 2026-27): sets, roster and set-builder form, empty, finite and equal sets, subsets and intervals, power set, union, intersection, complement and De Morgan's laws. All 49 questions are answered, with the key answer highlighted.

NCERT textbook Chapter 1 (PDF)

Exam revision? Chapter 1 notes with key points, diagrams and practice questions: see the free sample · ₹199 for all 14 chapters

Free NCERT solutions by Notes Bazar · www.notesbazar.in/ncert-solutions/class-11-maths/chapter-1-sets

Roster form lists the elements; set-builder form states the property. A set with n elements has $2^{n}$ subsets. $A \cup B$ , $A \cap B$ , $A - B$ , $A^{'} = U - A$ ; De Morgan: $(A \cup B)^{'} = A^{'} \cap B^{'}$ , $(A \cap B)^{'} = A^{'} \cup B^{'}$ . $n (A \cup B) = n (A) + n (B) - n (A \cap B)$ .

Exercise 1.1

Which of the following are sets? Justify. (i) Months of a year beginning with J (ii) Ten most talented writers of India (iii) A team of eleven best cricket batsmen of the world (iv) All boys in your class (v) All natural numbers less than 100 (vi) Novels written by Munshi Prem Chand (vii) All even integers (viii) The questions in this chapter (ix) The most dangerous animals of the world.

Solution

A collection is a set only if membership can be decided definitely. (i) {January, June, July}: set. (ii) "Most talented" depends on opinion: not a set. (iii) "Best" is a matter of opinion: not a set. (iv) set. (v) {1, 2, …, 99}: set. (vi) set (the novels are known). (vii) set. (viii) set. (ix) "Most dangerous" is not well defined: not a set. So (i), (iv), (v), (vi), (vii), (viii) are sets.

(i) set; (ii) not a set; (iii) not a set; (iv) set; (v) set; (vi) set; (vii) set; (viii) set; (ix) not a set

A = {1, 2, 3, 4, 5, 6}. Insert ∈ or ∉: (i) 5 … A (ii) 8 … A (iii) 0 … A (iv) 4 … A (v) 2 … A (vi) 10 … A.

Solution

(i) 5 ∈ A (ii) 8 ∉ A (iii) 0 ∉ A (iv) 4 ∈ A (v) 2 ∈ A (vi) 10 ∉ A.

(i) ∈; (ii) ∉; (iii) ∉; (iv) ∈; (v) ∈; (vi) ∉

Write in roster form: (i) A = {x : x an integer, −3 ≤ x < 7} (ii) B = {x : x a natural number less than 6} (iii) C = {x : x a two-digit natural number whose digits add to 8} (iv) D = {x : x a prime divisor of 60} (v) E = letters of TRIGONOMETRY (vi) F = letters of BETTER.

Solution

(i) {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6} (7 excluded). (ii) {1, 2, 3, 4, 5}. (iii) Tens digit 1 to 8 with units 8 − tens: {17, 26, 35, 44, 53, 62, 71, 80}. (iv) 60 = 2² × 3 × 5: {2, 3, 5}. (v) {T, R, I, G, O, N, M, E, Y} (each letter once). (vi) {B, E, T, R}.

(i) {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6}; (ii) {1, 2, 3, 4, 5}; (iii) {17, 26, 35, 44, 53, 62, 71, 80}; (iv) {2, 3, 5}; (v) {T, R, I, G, O, N, M, E, Y}; (vi) {B, E, T, R}

Write in set-builder form: (i) {3, 6, 9, 12} (ii) {2, 4, 8, 16, 32} (iii) {5, 25, 125, 625} (iv) {2, 4, 6, …} (v) {1, 4, 9, …, 100}.

Solution

(i) {x : x = 3n, n ∈ N, 1 ≤ n ≤ 4}. (ii) {x : x = 2ⁿ, n ∈ N, 1 ≤ n ≤ 5}. (iii) {x : x = 5ⁿ, n ∈ N, 1 ≤ n ≤ 4}. (iv) {x : x is an even natural number}. (v) {x : x = n², n ∈ N, 1 ≤ n ≤ 10}.

(i) {x : x = 3n, n ∈ N, 1 ≤ n ≤ 4}; (ii) {x : x = 2ⁿ, n ∈ N, 1 ≤ n ≤ 5}; (iii) {x : x = 5ⁿ, n ∈ N, 1 ≤ n ≤ 4}; (iv) {x : x is an even natural number}; (v) {x : x = n², n ∈ N, 1 ≤ n ≤ 10}

List the elements: (i) A = {x : x an odd natural number} (ii) B = {x : x an integer, −½ < x < 9/2} (iii) C = {x : x an integer, x² ≤ 4} (iv) D = letters of LOYAL (v) E = months not having 31 days (vi) F = consonants preceding k.

Solution

(i) {1, 3, 5, …}. (ii) Integers strictly between −0.5 and 4.5: {0, 1, 2, 3, 4}. (iii) x² ≤ 4 ⇒ −2 ≤ x ≤ 2: {−2, −1, 0, 1, 2}. (iv) {L, O, Y, A}. (v) {February, April, June, September, November}. (vi) {b, c, d, f, g, h, j}.

(i) {1, 3, 5, …}; (ii) {0, 1, 2, 3, 4}; (iii) {−2, −1, 0, 1, 2}; (iv) {L, O, Y, A}; (v) {February, April, June, September, November}; (vi) {b, c, d, f, g, h, j}

Match: (i) {1, 2, 3, 6} (ii) {2, 3} (iii) {M, A, T, H, E, I, C, S} (iv) {1, 3, 5, 7, 9} with (a) {x : x a prime divisor of 6} (b) {x : x an odd natural number less than 10} (c) {x : x a natural number and divisor of 6} (d) {x : x a letter of MATHEMATICS}.

Solution

(i) ↔ (c); (ii) ↔ (a); (iii) ↔ (d); (iv) ↔ (b).

(i) (c); (ii) (a); (iii) (d); (iv) (b)

Exercise 1.2

Which are null sets? (i) odd natural numbers divisible by 2 (ii) even primes (iii) {x : x ∈ N, x < 5 and x > 7} (iv) {y : y a point common to any two parallel lines}.

Solution

(i) No odd number is divisible by 2: null. (ii) {2}: not null. (iii) No number is both < 5 and > 7: null. (iv) Parallel lines never meet: null. Answer: (i), (iii), (iv).

(i) null; (ii) not null; (iii) null; (iv) null

Finite or infinite? (i) months of a year (ii) {1, 2, 3, …} (iii) {1, 2, …, 100} (iv) positive integers greater than 100 (v) primes less than 99.

Solution

(i) 12 elements: finite. (ii) infinite. (iii) 100 elements: finite. (iv) infinite. (v) finite (25 primes).

(i) finite; (ii) infinite; (iii) finite; (iv) infinite; (v) finite

Finite or infinite? (i) lines parallel to the x-axis (ii) letters of the English alphabet (iii) multiples of 5 (iv) animals living on the earth (v) circles through the origin.

Solution

(i) infinite (y = c for every real c). (ii) finite (26). (iii) infinite. (iv) finite (very large but definite). (v) infinite (any centre gives one).

(i) infinite; (ii) finite; (iii) infinite; (iv) finite; (v) infinite

Is A = B? (i) {a, b, c, d}, {d, c, b, a} (ii) {4, 8, 12, 16}, {8, 4, 16, 18} (iii) {2, 4, 6, 8, 10}, {x : x a positive even integer, x ≤ 10} (iv) {x : x a multiple of 10}, {10, 15, 20, 25, 30, …}.

Solution

(i) Same elements: A = B. (ii) 12 ∈ A but 12 ∉ B: A ≠ B. (iii) B = {2, 4, 6, 8, 10}: A = B. (iv) 15 ∈ B but 15 ∉ A: A ≠ B.

(i) A = B; (ii) A ≠ B; (iii) A = B; (iv) A ≠ B

Are these equal? (i) A = {2, 3}, B = {x : x a solution of x² + 5x + 6 = 0} (ii) A = letters of FOLLOW, B = letters of WOLF.

Solution

(i) x² + 5x + 6 = (x + 2)(x + 3) = 0 gives B = {−2, −3} ≠ {2, 3}: not equal. (ii) A = {F, O, L, W}, B = {W, O, L, F}: equal.

not equal; equal

Select equal sets: A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D = {3, 1, 4, 2}, E = {−1, 1}, F = {0, a}, G = {1, −1}, H = {0, 1}.

Solution

B = D (both {1, 2, 3, 4}) and E = G (both {−1, 1}); no other pair has the same elements.

B = D; E = G

Exercise 1.3

Fill ⊂ or ⊄: (i) {2, 3, 4} … {1, 2, 3, 4, 5} (ii) {a, b, c} … {b, c, d} (iii) {x : x a student of Class XI of your school} … {x : x a student of your school} (iv) {x : x a circle in the plane} … {x : x a circle in the same plane with radius 1 unit} (v) {x : x a triangle in a plane} … {x : x a rectangle in the plane} (vi) {x : x an equilateral triangle in a plane} … {x : x a triangle in the same plane} (vii) {x : x an even natural number} … {x : x an integer}.

Solution

(i) ⊂. (ii) a is not in the second: ⊄. (iii) ⊂. (iv) A circle of radius 2 is not in the second: ⊄. (v) A triangle is not a rectangle: ⊄. (vi) ⊂. (vii) ⊂.

(i) ⊂; (ii) ⊄; (iii) ⊂; (iv) ⊄; (v) ⊄; (vi) ⊂; (vii) ⊂

True or false? (i) {a, b} ⊄ {b, c, a} (ii) {a, e} ⊂ {x : x a vowel of the English alphabet} (iii) {1, 2, 3} ⊂ {1, 3, 5} (iv) {a} ⊂ {a, b, c} (v) {a} ∈ {a, b, c} (vi) {x : x an even natural number less than 6} ⊂ {x : x a natural number which divides 36}.

Solution

(i) a, b are both in {b, c, a}, so {a, b} ⊂ it: False. (ii) a, e are vowels: True. (iii) 2 ∉ {1, 3, 5}: False. (iv) True. (v) The elements are a, b, c, not {a}: False. (vi) {2, 4} and both divide 36: True.

(i) False; (ii) True; (iii) False; (iv) True; (v) False; (vi) True

A = {1, 2, {3, 4}, 5}. Which are incorrect and why? (i) {3, 4} ⊂ A (ii) {3, 4} ∈ A (iii) {{3, 4}} ⊂ A (iv) 1 ∈ A (v) 1 ⊂ A (vi) {1, 2, 5} ⊂ A (vii) {1, 2, 5} ∈ A (viii) {1, 2, 3} ⊂ A (ix) φ ∈ A (x) φ ⊂ A (xi) {φ} ⊂ A.

Solution

The elements of A are 1, 2, {3, 4} and 5. Incorrect: (i) 3 ∉ A (the element is {3, 4}, not 3); (v) 1 is an element, not a set (correct: {1} ⊂ A); (vii) {1, 2, 5} is not an element of A (it is a subset); (viii) 3 ∉ A; (ix) φ is not an element of A; (xi) φ ∉ A, so {φ} ⊄ A. Correct: (ii), (iii), (iv), (vi), (x).

(xi) Correct

Write all subsets of (i) {a} (ii) {a, b} (iii) {1, 2, 3} (iv) φ.

Solution

(i) φ, {a}. (ii) φ, {a}, {b}, {a, b}. (iii) φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3} (2³ = 8). (iv) φ only.

(i) φ, {a}; (ii) φ, {a}, {b}, {a, b}; (iii) φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}; (iv) φ

Write as intervals: (i) {x : x ∈ R, −4 < x ≤ 6} (ii) {x : x ∈ R, −12 < x < −10} (iii) {x : x ∈ R, 0 ≤ x < 7} (iv) {x : x ∈ R, 3 ≤ x ≤ 4}.

Solution

(i) (−4, 6] (ii) (−12, −10) (iii) [0, 7) (iv) [3, 4].

(i) (−4, 6]; (ii) (−12, −10); (iii) [0, 7); (iv) [3, 4]

Write in set-builder form: (i) (−3, 0) (ii) [6, 12] (iii) (6, 12] (iv) [−23, 5).

Solution

(i) {x : x ∈ R, −3 < x < 0} (ii) {x : x ∈ R, 6 ≤ x ≤ 12} (iii) {x : x ∈ R, 6 < x ≤ 12} (iv) {x : x ∈ R, −23 ≤ x < 5}.

(i) {x : x ∈ R, −3 < x < 0}; (ii) {x : x ∈ R, 6 ≤ x ≤ 12}; (iii) {x : x ∈ R, 6 < x ≤ 12}; (iv) {x : x ∈ R, −23 ≤ x < 5}

What universal set would you propose for (i) the set of right triangles (ii) the set of isosceles triangles?

Solution

For both: the set of all triangles (in a plane); the set of all polygons would also do.

the set of all triangles

A = {1, 3, 5}, B = {2, 4, 6}, C = {0, 2, 4, 6, 8}. Which can be a universal set for all three? (i) {0, 1, 2, 3, 4, 5, 6} (ii) φ (iii) {0, 1, 2, …, 10} (iv) {1, 2, …, 8}.

Solution

U must contain A ∪ B ∪ C = {0, 1, 2, 3, 4, 5, 6, 8}. (i) lacks 8; (ii) empty; (iv) lacks 0. Only (iii).

(iv) (iii)

Exercise 1.4

Find the union: (i) X = {1, 3, 5}, Y = {1, 2, 3} (ii) A = {a, e, i, o, u}, B = {a, b, c} (iii) A = {x : x a natural multiple of 3}, B = {x : x a natural number less than 6} (iv) A = {x : x ∈ N, 1 < x ≤ 6}, B = {x : x ∈ N, 6 < x < 10} (v) A = {1, 2, 3}, B = φ.

Solution

(i) {1, 2, 3, 5}. (ii) {a, b, c, e, i, o, u}. (iii) A = {3, 6, 9, …}, B = {1, 2, 3, 4, 5}: A ∪ B = {1, 2, 4, 5} together with all multiples of 3 = {x : x = 1, 2, 4, 5 or a multiple of 3}. (iv) {2, …, 6} ∪ {7, 8, 9} = {x : x ∈ N, 1 < x < 10}. (v) {1, 2, 3}.

(i) {1, 2, 3, 5}; (ii) {a, b, c, e, i, o, u}; (iii) {1, 2, 4, 5} together with all multiples of 3; (iv) {x : x ∈ N, 1 < x < 10}; (v) {1, 2, 3}

A = {a, b}, B = {a, b, c}. Is A ⊂ B? What is A ∪ B?

Solution

Yes, A ⊂ B; A ∪ B = {a, b, c} = B.

Yes; {a, b, c}

If A ⊂ B, what is A ∪ B?

Solution

Every element of A is already in B, so A ∪ B = B.

A ∪ B = B

A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8}, D = {7, 8, 9, 10}. Find (i) A ∪ B (ii) A ∪ C (iii) B ∪ C (iv) B ∪ D (v) A ∪ B ∪ C (vi) A ∪ B ∪ D (vii) B ∪ C ∪ D.

Solution

(i) {1, 2, 3, 4, 5, 6} (ii) {1, 2, 3, 4, 5, 6, 7, 8} (iii) {3, 4, 5, 6, 7, 8} (iv) {3, 4, 5, 6, 7, 8, 9, 10} (v) {1, 2, …, 8} (vi) {1, 2, …, 10} (vii) {3, 4, …, 10}.

(i) {1, 2, 3, 4, 5, 6}; (ii) {1, 2, 3, 4, 5, 6, 7, 8}; (iii) {3, 4, 5, 6, 7, 8}; (iv) {3, 4, 5, 6, 7, 8, 9, 10}; (v) {1, 2, …, 8}; (vi) {1, 2, …, 10}; (vii) {3, 4, …, 10}

Find the intersection of each pair of question 1.

Solution

(i) {1, 3} (ii) {a} (iii) {3, 6, …} ∩ {1, …, 5} = {3} (iv) φ (v) φ.

(i) {1, 3}; (ii) {a}; (iii) {3}; (iv) φ; (v) φ

A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15}, D = {15, 17}. Find (i) A ∩ B (ii) B ∩ C (iii) A ∩ C ∩ D (iv) A ∩ C (v) B ∩ D (vi) A ∩ (B ∪ C) (vii) A ∩ D (viii) A ∩ (B ∪ D) (ix) (A ∩ B) ∩ (B ∪ C) (x) (A ∪ D) ∩ (B ∪ C).

Solution

(i) {7, 9, 11} (ii) {11, 13} (iii) A ∩ C = {11}, then ∩ D: φ (iv) {11} (v) φ (vi) B ∪ C = {7, 9, 11, 13, 15}: {7, 9, 11} (vii) φ (viii) B ∪ D = {7, 9, 11, 13, 15, 17}: {7, 9, 11} (ix) {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11} (x) {3, 5, 7, 9, 11, 15, 17} ∩ {7, 9, 11, 13, 15} = {7, 9, 11, 15}.

(i) {7, 9, 11}; (ii) {11, 13}; (iii) φ; (iv) {11}; (v) φ; (vi) {7, 9, 11}; (vii) φ; (viii) {7, 9, 11}; (ix) {7, 9, 11}; (x) {7, 9, 11, 15}

A = N, B = even naturals, C = odd naturals, D = primes. Find (i) A ∩ B (ii) A ∩ C (iii) A ∩ D (iv) B ∩ C (v) B ∩ D (vi) C ∩ D.

Solution

(i) B (even naturals) (ii) C (iii) D (iv) no number is both even and odd: φ (v) the only even prime: {2} (vi) {x : x an odd prime}.

(i) B; (ii) C; (iii) D; (iv) φ; (v) {2}; (vi) {x : x an odd prime}

Which pairs are disjoint? (i) {1, 2, 3, 4} and {x : x ∈ N, 4 ≤ x ≤ 6} (ii) {a, e, i, o, u} and {c, d, e, f} (iii) {x : x an even integer} and {x : x an odd integer}.

Solution

(i) 4 is common: not disjoint. (ii) e is common: not disjoint. (iii) No integer is both: disjoint.

(iii) disjoint

A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16}, D = {5, 10, 15, 20}. Find (i) A − B (ii) A − C (iii) A − D (iv) B − A (v) C − A (vi) D − A (vii) B − C (viii) B − D (ix) C − B (x) D − B (xi) C − D (xii) D − C.

Solution

(i) {3, 6, 9, 15, 18, 21} (ii) {3, 9, 15, 18, 21} (iii) {3, 6, 9, 12, 18, 21} (iv) {4, 8, 16, 20} (v) {2, 4, 8, 10, 14, 16} (vi) {5, 10, 20} (vii) {20} (viii) {4, 8, 12, 16} (ix) {2, 6, 10, 14} (x) {5, 10, 15} (xi) {2, 4, 6, 8, 12, 14, 16} (xii) {5, 15, 20}.

(i) {3, 6, 9, 15, 18, 21}; (ii) {3, 9, 15, 18, 21}; (iii) {3, 6, 9, 12, 18, 21}; (iv) {4, 8, 16, 20}; (v) {2, 4, 8, 10, 14, 16}; (vi) {5, 10, 20}; (vii) {20}; (viii) {4, 8, 12, 16}; (ix) {2, 6, 10, 14}; (x) {5, 10, 15}; (xi) {2, 4, 6, 8, 12, 14, 16}; (xii) {5, 15, 20}

X = {a, b, c, d}, Y = {f, b, d, g}. Find (i) X − Y (ii) Y − X (iii) X ∩ Y.

Solution

(i) {a, c} (ii) {f, g} (iii) {b, d}.

(i) {a, c}; (ii) {f, g}; (iii) {b, d}

If R is the set of real numbers and Q the set of rational numbers, what is R − Q?

Solution

The real numbers that are not rational: the set of irrational numbers.

the set of irrational numbers

True or false? Justify. (i) {2, 3, 4, 5} and {3, 6} are disjoint (ii) {a, e, i, o, u} and {a, b, c, d} are disjoint (iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint (iv) {2, 6, 10} and {3, 7, 11} are disjoint.

Solution

(i) False: 3 is common. (ii) False: a is common. (iii) True: the intersection is φ. (iv) True: the intersection is φ.

(i) False; (ii) False; (iii) True; (iv) True

Exercise 1.5

U = {1, 2, …, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8}, C = {3, 4, 5, 6}. Find (i) A′ (ii) B′ (iii) (A ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ (vi) (B − C)′.

Solution

(i) {5, 6, 7, 8, 9} (ii) {1, 3, 5, 7, 9} (iii) A ∪ C = {1, …, 6}: {7, 8, 9} (iv) A ∪ B = {1, 2, 3, 4, 6, 8}: {5, 7, 9} (v) A = {1, 2, 3, 4} (vi) B − C = {2, 8}: {1, 3, 4, 5, 6, 7, 9}.

(i) {5, 6, 7, 8, 9}; (ii) {1, 3, 5, 7, 9}; (iii) {7, 8, 9}; (iv) {5, 7, 9}; (v) A = {1, 2, 3, 4}; (vi) {1, 3, 4, 5, 6, 7, 9}

U = {a, b, c, d, e, f, g, h}. Find the complements of (i) {a, b, c} (ii) {d, e, f, g} (iii) {a, c, e, g} (iv) {f, g, h, a}.

Solution

(i) {d, e, f, g, h} (ii) {a, b, c, h} (iii) {b, d, f, h} (iv) {b, c, d, e}.

(i) {d, e, f, g, h}; (ii) {a, b, c, h}; (iii) {b, d, f, h}; (iv) {b, c, d, e}

With U = N, write the complements of (i) even naturals (ii) odd naturals (iii) positive multiples of 3 (iv) primes (v) naturals divisible by 3 and 5 (vi) perfect squares (vii) perfect cubes (viii) {x : x + 5 = 8} (ix) {x : 2x + 5 = 9} (x) {x : x ≥ 7} (xi) {x : x ∈ N, 2x + 1 > 10}.

Solution

(i) odd naturals (ii) even naturals (iii) {x : x ∈ N, x not a multiple of 3} (iv) 1 and the composite numbers (v) {x : x ∈ N, x not divisible by 15} (i.e. not divisible by 3 or not by 5) (vi) naturals that are not perfect squares (vii) not perfect cubes (viii) {3}′ = N − {3} = {x : x ∈ N, x ≠ 3} (ix) {2}′ = N − {2} (x) {1, 2, 3, 4, 5, 6} (xi) 2x + 1 > 10 ⇔ x ≥ 5, so the complement is {1, 2, 3, 4} = {x : x ∈ N, x ≤ 9/2}.

(i) odd naturals; (ii) even naturals; (iii) {x : x ∈ N, x not a multiple of 3}; (iv) 1 and the composite numbers; (v) {x : x ∈ N, x not divisible by 15}; (vi) naturals that are not perfect squares; (vii) not perfect cubes; (viii) N − {3}; (ix) N − {2}; (x) {1, 2, 3, 4, 5, 6}; (xi) {1, 2, 3, 4}

U = {1, …, 9}, A = {2, 4, 6, 8}, B = {2, 3, 5, 7}. Verify (i) (A ∪ B)′ = A′ ∩ B′ (ii) (A ∩ B)′ = A′ ∪ B′.

Solution

A′ = {1, 3, 5, 7, 9}, B′ = {1, 4, 6, 8, 9}. (i) A ∪ B = {2, 3, 4, 5, 6, 7, 8}, (A ∪ B)′ = {1, 9}; A′ ∩ B′ = {1, 9}: equal. (ii) A ∩ B = {2}, (A ∩ B)′ = {1, 3, 4, 5, 6, 7, 8, 9}; A′ ∪ B′ = {1, 3, 4, 5, 6, 7, 8, 9}: equal.

(i) {1, 9}, {1, 9}; (ii) {1, 3, 4, 5, 6, 7, 8, 9}, {1, 3, 4, 5, 6, 7, 8, 9}

Draw Venn diagrams for (i) (A ∪ B)′ (ii) A′ ∩ B′ (iii) (A ∩ B)′ (iv) A′ ∪ B′.

Solution

Draw U as a rectangle with two overlapping circles A and B. (i) and (ii): shade everything outside both circles (the two diagrams are the same: De Morgan). (iii) and (iv): shade everything except the lens-shaped overlap A ∩ B (again identical).

(i) everything outside both circles; (iii) everything except the lens-shaped overlap

U = all triangles in a plane. If A is the set of triangles with at least one angle different from 60°, what is A′?

Solution

A′ = triangles with no angle different from 60°, i.e. all three angles 60°: the set of equilateral triangles.

the set of equilateral triangles

Fill in: (i) A ∪ A′ = … (ii) φ′ ∩ A = … (iii) A ∩ A′ = … (iv) U′ ∩ A = ….

Solution

(i) U (ii) φ′ = U, so U ∩ A = A (iii) φ (iv) U′ = φ, so φ.

(i) U; (ii) A; (iii) φ; (iv) φ

Miscellaneous Exercise on Chapter 1

Decide which of A = {x : x ∈ R, x² − 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8, …}, D = {6} are subsets of one another.

Solution

x² − 8x + 12 = (x − 2)(x − 6) = 0: A = {2, 6}. Hence D ⊂ A ⊂ B ⊂ C (so also D ⊂ B, D ⊂ C, A ⊂ C).

D ⊂ A ⊂ B ⊂ C

True or false? Prove if true, give an example if false. (i) x ∈ A, A ∈ B ⇒ x ∈ B (ii) A ⊂ B, B ∈ C ⇒ A ∈ C (iii) A ⊂ B, B ⊂ C ⇒ A ⊂ C (iv) A ⊄ B, B ⊄ C ⇒ A ⊄ C (v) x ∈ A, A ⊄ B ⇒ x ∈ B (vi) A ⊂ B, x ∉ B ⇒ x ∉ A.

Solution

(i) False: A = {1}, B = {{1}, 2}: 1 ∈ A, A ∈ B, but 1 ∉ B. (ii) False: A = {1}, B = {1, 2}, C = {{1, 2}, 3}: A ⊂ B, B ∈ C, but A ∉ C. (iii) True: x ∈ A ⇒ x ∈ B (A ⊂ B) ⇒ x ∈ C (B ⊂ C); so A ⊂ C. (iv) False: A = {1, 2}, B = {2, 3}, C = {1, 2, 4}: A ⊄ B, B ⊄ C, yet A ⊂ C. (v) False: A = {1, 2}, B = {2, 3}: 1 ∈ A, A ⊄ B, but 1 ∉ B. (vi) True: if x ∈ A then x ∈ B (A ⊂ B), contradicting x ∉ B; so x ∉ A.

(i) False; (ii) False; (iii) True; (iv) False; (v) False; (vi) True

If A ∪ B = A ∪ C and A ∩ B = A ∩ C, show that B = C.

Solution

B = B ∩ (A ∪ B) = B ∩ (A ∪ C) = (B ∩ A) ∪ (B ∩ C) (distributive) = (A ∩ C) ∪ (B ∩ C) (since A ∩ B = A ∩ C) = (A ∪ B) ∩ C = (A ∪ C) ∩ C = C. (The first step uses B ⊂ A ∪ B.)

Show that the following four conditions are equivalent: (i) A ⊂ B (ii) A − B = φ (iii) A ∪ B = B (iv) A ∩ B = A.

Solution

(i) ⇒ (ii): every x ∈ A is in B, so no element is in A but not in B: A − B = φ. (ii) ⇒ (iii): A − B = φ means every x ∈ A lies in B, so A ∪ B ⊂ B; and B ⊂ A ∪ B always: A ∪ B = B. (iii) ⇒ (iv): A ⊂ A ∪ B = B, so every x ∈ A is in A ∩ B: A ⊂ A ∩ B ⊂ A, hence A ∩ B = A. (iv) ⇒ (i): A = A ∩ B ⊂ B. So all four are equivalent.

(iv) equivalent

Show that if A ⊂ B, then C − B ⊂ C − A.

Solution

Let x ∈ C − B: x ∈ C and x ∉ B. Since A ⊂ B, x ∉ B ⇒ x ∉ A. So x ∈ C and x ∉ A: x ∈ C − A. Hence C − B ⊂ C − A.

C − B ⊂ C − A

Show that for any sets A and B, A = (A ∩ B) ∪ (A − B) and A ∪ (B − A) = A ∪ B.

Solution

(A ∩ B) ∪ (A − B) = (A ∩ B) ∪ (A ∩ B′) = A ∩ (B ∪ B′) = A ∩ U = A. A ∪ (B − A) = A ∪ (B ∩ A′) = (A ∪ B) ∩ (A ∪ A′) = (A ∪ B) ∩ U = A ∪ B.

A; A ∪ B

Using properties of sets, show (i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A.

Solution

(i) A ∪ (A ∩ B) = (A ∩ U) ∪ (A ∩ B) = A ∩ (U ∪ B) = A ∩ U = A. (ii) A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B) = A ∪ (A ∩ B) = A by (i).

Show that A ∩ B = A ∩ C need not imply B = C.

Solution

Take A = {1, 2}, B = {1, 3}, C = {1, 4}: A ∩ B = {1} = A ∩ C, but B ≠ C.

B ≠ C

If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.

Solution

A = A ∩ (A ∪ X) = A ∩ (B ∪ X) = (A ∩ B) ∪ (A ∩ X) = (A ∩ B) ∪ φ = A ∩ B. Similarly B = B ∩ (B ∪ X) = B ∩ (A ∪ X) = (B ∩ A) ∪ (B ∩ X) = A ∩ B. Hence A = B.

A = B

Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ.

Solution

A = {1, 2}, B = {2, 3}, C = {1, 3}: A ∩ B = {2}, B ∩ C = {3}, A ∩ C = {1} (all non-empty), and A ∩ B ∩ C = φ.

A ∩ B ∩ C = φ

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